XY is a line parallel to side BC of triangle ABC BE ll AC and CF ll ab - Maths - Quadrilaterals

Question

XY is a line parallel to side BC of triangle ABC BE ll AC and CF ll
ab meets xy at e and f respectively show that ar(AEB)=ar(acf)

Abhishek Jha · Accepted Answer

Dear Student, Here is the solution of your asked query:

It is given that XY || BC ⇒ EY || BC BE || AC ⇒ BE || CY Therefore, EBCY is a parallelogram It is given that XY || BC ⇒ XF || BC FC || AB ⇒ FC || XB Therefore, BCFX is a parallelogram Parallelograms EBCY and BCFX are on the same base BC and between the same parallels BC and EF ∴ Area (EBCY) = Area (BCFX) (1) Consider parallelogram EBCY and ΔAEB These lie on the same base BE and are between the same parallels BE and AC ∴ Area (ΔABE) =

Area (EBCY) (2) Also, parallelogram BCFX and ΔACF are on the same base CF and between the same parallels CF and AB ∴ Area (ΔACF) =

Area (BCFX) (3) From equations (1), (2), and (3), we obtain Area (ΔABE) = Area (ΔACF) Regards

XY is a line parallel to side BC of triangle ABC BE ll AC and CF ll ab meets xy at e and f respectively . show that ar(AEB)=ar(acf)